Problem: Find $\tan\left(195^\circ\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1-\sqrt{6}}{1+\sqrt{6}}$ (Choice B) B $\dfrac{3+\sqrt{3}}{3-\sqrt{3}}$ (Choice C) C $\dfrac{1-\sqrt{3}}{2}$ (Choice D) D $\dfrac{3-\sqrt{3}}{3+\sqrt{3}}$
The strategy First, we should rewrite the given angle $195^\circ$ as the sum or difference of two special angles. Then, we can use the tangent addition or subtraction identities in order to evaluate $\tan\left(195^\circ\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $195^\circ$ We can rewrite $195^\circ$ as follows. $\begin{aligned}195^\circ&=225^\circ-30^\circ\end{aligned}$ In other words, $195^\circ$ is the difference of the special angles $225^\circ$ and $30^\circ$. Evaluating $\tan\left(195^\circ\right)$ Using the tangent subtraction identity, we get the following. $\begin{aligned} \tan\left(195^\circ\right)&= \tan\left(225^\circ-30^\circ\right) \\\\\\ &= \dfrac{\tan\left(225^\circ\right)-\tan\left(30^\circ\right)}{1+\tan\left(225^\circ\right)\tan\left(30^\circ\right)}\\\\\\ &= \dfrac{\left(1\right)-\left(\dfrac{\sqrt{3}}{3}\right)}{1+\left(1\right)\left(\dfrac{\sqrt{3}}{3}\right)}\\\\\\ &=\dfrac{3-\sqrt{3}}{3+\sqrt{3}} \end{aligned}$ Summary $\tan\left(195^\circ\right) = \dfrac{3-\sqrt{3}}{3+\sqrt{3}}$